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Published byJames Griffin Modified over 9 years ago
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Coulomb’s Law Point Charge :
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Line Charge :
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Surface Charge :
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Volume Charge
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Prob. 2.6: Electric Field at a height z above the centre of a circular plate of uniform charge density.
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Limiting Case : (Infinite Sheet)
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Prob. 2.41. Electric Field at a height z above the centre of a square plate of uniform charge density.
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Any kind of charge distribution can be treated as a volume distribution 1. Point charges :
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2. Line Charge
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3. Surface Charge
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Divergence of the Electric Field
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Examples : 1. Point Charge
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Infinite Charge Sheet
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Gauss’ Law
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Applications of Gauss’ Law Although Gauss’ Law is valid for any kind of charge distribution and any Gaussian Surface, its applicability to determine the electric field is restricted only to symmetrical charge distribution
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1. Electric field of a point charge Gaussian Surface
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Prob. 2.16 Long co-axial cable : i) Inner solid cylinder of radius a carrying uniform volume charge density ρ ii) outer cylindrical surface of radius b carrying equal and opposite charge of uniform density σ. Find field in regions i) s b
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s < a
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a < s < b
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s > a
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Prob. 2.17 Infinite plane slab of thickness 2d (-d d
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-d < y < d
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Curl of the Electric Field
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Where, V can be constructed as :
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The scalar field is called the electric potential and the point is the zero of the potential
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Changing the zero of the potential Let be the potentials with the zero (reference points) at respectively
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Prob. 2.20 One of the following is an impossible electrostatic field. Which one? For the possible one, find the potential and show that it gives the correct field
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Conventionally, the zero of the potential is taken at infinity : Potential of a point charge (zero at infinity) :
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Charge located at :
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Potentials of Extended Charge Distributions : Line Charge : Surface Charge :
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Volume Charge : Prob. 2.26 The conical surface has uniform charge density σ. Find p.d between points a & b
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Prob. 2.9 Suppose the electric field in some region is found to be : a) Find the charge distribution that could produce this field b) Find the total charge contained in a sphere of radius R centered on the origin. Do it in two different ways.
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Work done in moving a charge in an electric field If the charge is brought from infinity to the point :
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If V is the potential of a point charge Q located at then,
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Electrostatic Energy of a Charge Distribution It is the work done to assemble the charge configuration, starting from some initial configuration Initial Config. Given Config.
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The standard initial configuration is taken to be one in which all small (infinitesimal) pieces of charge are infinitely separated from one another.
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1. Point Charges
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Electrostatic Energy of Continuous Charge Distribution :
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The electrostatic energy of a charge distribution can be expressed as an integral over the electric field of the distribution :
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Proof :
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Including more and more volume in the integral,
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Prob. 2.45 A sphere of radius R carries a charge density. Find the energy of the configuration in two different ways. a) Find the energy by integrating over the field b) Find the potential everywhere and do the integral :
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Prob. 2.33 Find the electrostatic energy of a uniformly charged solid sphere of total charge Q by the following method : Calculate work done in adding charge layer by layer
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Electrostatic energy of a point charge Q Energy of a uniform solid sphere of radius R and total charge Q : Energy of a point charge Q :
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Self and Interaction Energy are the fields produced by
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are the energies of the two charge distributions, existing alone. They are called the self energies of the distributions is the energy of interaction between them. It is the work done to bring them, already made, from infinity.
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Electrostatic Boundary Conditions 1 2 Applying Gauss’ law to the pillbox : As the two flat faces come infinitesimally close to the charged surface,
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Taking the line integral of around the closed loop : Combining (1) & (2) :
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Examples : 1 2 1. Infinite Sheet
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2. Co-axial Cable :
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Conductors A perfect conductor is a body possessing unlimited supply of charges of each kind (+ve & -ve), at least one of which kind is completely free to move within the body and on its surface. Mathematically a conductor is capable of developing any charge density with the only constraint : (Neutral Cond.)
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Or, since charge can reside only on the surface of a conductor, the only restriction on the surface charge density is : + + + + + + + + + + + + + + + + + ++ ++ - - - - - - - - - - - - - - - - - - - - -
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Properties of a perfect conductor 1. The electric field within the body of the conductor is zero At equilibrium (After charge flow in the conductor has ceased) : Otherwise, there is no reason why charge should stop flowing
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+ + + + + + + + + + + + + + + + + ++ ++ - - - - - - - - - - - - - - - - - - - - - i) Does the conductor have the necessary ammunition to nullify the external field within? ii) How long does it take the conductor to nullify the external field?
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2. There cannot be any charge density within the body of the conductor Gaussian surface S
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3. The surface of a conductor is an equipotential surface
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4. The electric field just outside the surface of a conductor is everywhere perpendicular to the surface Reason : The gradient is everywhere perpendicular the level surface.
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5. The electric field at any point just outside the surface of a conductor is related to the surface charge density at that point by : Reason : From boundary condition on the field :
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Electrostatic Pressure or
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Answer : Note : The surface of the conductor is everywhere pushed outwards.
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Correctly stated :
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Reason that
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Electrostatic Pressure :
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Prob. 2.38 A metal sphere of radius R carries a total charge Q. What is the force of repulsion between the two halves of the sphere?
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Poisson and Laplace Equation Combining, In a charge free region :
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Example (Point Charge) : However,
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Prob. 2.46 : The electric potential of some charge configuration is given by : Find the charge density and the total charge Q
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Ans:
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